228 CHAPTER 4 Polynomial and Rational Functions In Problems 81–86, find the bounds to the zeros of each polynomial function. Use the bounds to obtain a complete graph of f. 81. f x x x 2 1 3 2 ( ) = + − 82. f x x x x 3 2 4 3 2 ( ) = − + + 83. f x x x x 5 11 11 3 2 ( ) = − − + 84. f x x x x 2 11 6 3 2 ( ) = − − − 85. f x x x x 3 5 9 4 3 2 ( ) = + − + 86. f x x x x x 4 12 27 54 81 4 3 2 ( ) = − + − + In Problems 87–92, use the Intermediate Value Theorem to show that each function has a zero in the given interval. Approximate the zero correct to two decimal places. 87. f x x x x 8 2 5 1; 0, 1 4 2 ( ) [ ] = − + − 88. f x x x x 8 2; 1, 0 4 3 2 ( ) [ ] = + − + − 89. f x x x x 2 6 8 2; 5, 4 3 2 ( ) [ ] = + − + − − 90. f x x x 3 10 9; 3, 2 3 ( ) [ ] = − + − − 91. f x x x x x x 7 7 18 18; 1.4, 1.5 5 4 3 2 ( ) [ ] = − + − − + 92. f x x x x x x 3 2 6 2; 1.7, 1.8 5 4 3 2 ( ) [ ] = − − + + + Mixed Practice In Problems 93–104, graph each polynomial function by first factoring f and then analyzing f using Steps 1 through 8 on page 207. 93. f x x x x 2 5 6 3 2 ( ) = + − − 94. f x x x x 8 11 20 3 2 ( ) = + + − 95. f x x x x 2 2 1 3 2 ( ) = − + − 96. f x x x x 2 2 1 3 2 ( ) = + + + 97. f x x x 2 4 2 ( ) = + − 98. f x x x3 4 4 2 ( ) = − − 99. f x x x 4 7 2 4 2 ( ) = + − 100. f x x x 4 15 4 4 2 ( ) = + − 101. f x x x x x 3 2 4 3 2 ( ) = + − − + 102. f x x x x x 6 4 8 4 3 2 ( ) = − − + + 103. f x x x x 4 8 2 5 4 ( ) = − − + 104. f x x x x 4 12 3 5 4 ( ) = + − − 105. Mixed Practice Consider the function f x x x x x 3 16 24 192 6 4 3 2 ( ) = − − + + (a) Graph the polynomial function using a graphing utility. (b) Use the graphing utility to find the turning points. (c) Given that the first derivative of f is f x x x x 12 48 48 192, 3 2 ( ) ′ = − − + solve ( ) ′ = f x 0. Compare these solutions to the x-coordinates of the turning points from part (b). 106. Mixed Practice Consider the function f x x x x x 4 2 12 1 4 3 2 ( ) = − − + + − (a) Graph the polynomial function using a graphing utility. (b) Use the graphing utility to find the turning points. (c) Given that the first derivative of f is f x x x x 4 12 4 12, 3 2 ( ) ′ = − − + + solve f x 0 ( ) ′ = . Compare these solutions to the x-coordinates of the turning points from part (b). Applications and Extensions 107. Suppose that f x x x x 3 16 3 10 3 2 ( ) = + + − . Find the zeros of f x 3 ( ) + . 108. Suppose that f x x x x 4 11 26 24 3 2 ( ) = − − + . Find the zeros of f x 2 ( ) − . 109. Find k so that x 2 − is a factor of f x x kx kx 2 3 2 ( ) = − + + 110. Find k so that x 2 + is a factor of f x x kx kx 1 4 3 2 ( ) = − + + 111. What is the remainder when f x x x x 2 8 2 20 10 ( ) = − + − is divided by x 1? − 112. What is the remainder when f x x x x x 3 2 17 9 5 ( ) = − + − + is divided by x 1? + 113. Use the Factor Theorem to prove that x c − is a factor of x c n n − for any positive integer n. 114. Use the Factor Theorem to prove that x c + is a factor of x c n n + if n 1 ≥ is an odd integer. 115. One solution of the equation x x x 8 16 3 0 3 2 − + − = is 3. Find the sum of the remaining solutions. 116. One solution of the equation x x x 5 5 2 0 3 2 + + − = is 2− . Find the sum of the remaining solutions. 117. Geometry What is the length of the edge of a cube if, after a slice 1-inch thick is cut from one side, the volume remaining is 294 cubic inches? 118. Geometry What is the length of the edge of a cube if its volume is doubled by an increase of 6 centimeters in one edge, an increase of 12 centimeters in a second edge, and a decrease of 4 centimeters in the third edge? 119. Let f x( ) be a polynomial function whose coefficients are integers. Suppose that r is a real zero of f and that the leading coefficient of f is 1. Use the Rational Zeros Theorem to show that r is either an integer or an irrational number. 120. Bisection Method for Approximating Real Zeros of a Polynomial Function We begin with two consecutive integers, a and a 1, + for which f a( ) and f a 1 ( ) + are of opposite sign. Evaluate f at the midpoint m1 of a and a 1 + . If f m 0, 1 ( ) = then m1 is the zero of f, and we are finished. Otherwise, f m1 ( ) is of opposite sign to either f a( ) or f a 1 ( ) + . Suppose that it is f a( ) and f m1 ( ) that are of opposite sign. Now evaluate f at the midpoint m2 of a and m1. Repeat this process until the desired degree of accuracy is obtained. Note that each iteration places the zero in an interval whose length is half that of the previous interval. Use the bisection method to approximate the zero of f x x x x 8 2 5 1 4 2 ( ) = − + − in the interval 0, 1 [ ] correct to three decimal places. [Hint: The process ends when both endpoints agree to the desired number of decimal places.] 121. Challenge Problem Suppose f is a polynomial function. If f 2 7 ( ) − = and f 6 1, ( ) = − then the Intermediate Value Theorem guarantees which of the following? Justify your answer. (a) f 0 0 ( ) = (b) f c 3 ( ) = for at least one number c between 2− and 6. (c) f c 0 ( ) = for at least one number between 1− and 7. (d) f x 1 7 ( ) − ≤ ≤ for all numbers in the closed interval 2, 6 [ ] − .
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